# LeetCode: 1382. Balance a Binary Search Tree

## 题目¶

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

```Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

0
/ \
-3   9
/   /
-10  5
```

## 解法¶

### 先得到节点组成的有序数组，然后将有序数组转换为平衡二叉搜索树¶

• 将数组从中间切分，
• 中间那个数是根节点，
• 左边的数组是左子节点构成的数组，
• 右边的数组是右子节点构成的数组，
• 按同样的方法切分左边数组和右边数组。

```def inorder(root):
if root is None:
return
inorder(root.left)

# print(root)

inorder(root.right)
```

```# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
def balanceBST(self, root):
self.nodes = []
self.inorder(root)
root = self.buildBST(self.nodes, 0, len(self.nodes) - 1)
return root

def buildBST(self, nodes, min_index, max_index):
if min_index > max_index:
return

mid_index = min_index + (max_index - min_index)/2
root = nodes[mid_index]
root.left = self.buildBST(nodes, min_index, mid_index - 1)
root.right = self.buildBST(nodes, mid_index + 1, max_index)

return root

def inorder(self, root):
if root is None:
return
self.inorder(root.left)

self.nodes.append(root)

self.inorder(root.right)
```

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